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Oracle Java SE 21 Developer Professional Sample Questions (Q10-Q15):

NEW QUESTION # 10
Given:
java
List<Long> cannesFestivalfeatureFilms = LongStream.range(1, 1945)
.boxed()
.toList();
try (var executor = Executors.newVirtualThreadPerTaskExecutor()) {
cannesFestivalfeatureFilms.stream()
.limit(25)
.forEach(film -> executor.submit(() -> {
System.out.println(film);
}));
}
What is printed?

A. Numbers from 1 to 1945 randomly
B. Numbers from 1 to 25 randomly
C. Compilation fails
D. An exception is thrown at runtime
E. Numbers from 1 to 25 sequentially

Answer: B

Explanation:
* Understanding LongStream.range(1, 1945).boxed().toList();
* LongStream.range(1, 1945) generates a stream of numbersfrom 1 to 1944.
* .boxed() converts the primitive long values to Long objects.
* .toList() (introduced in Java 16)creates an immutable list.
* Understanding Executors.newVirtualThreadPerTaskExecutor()
* Java 21 introducedvirtual threadsto improve concurrency.
* Executors.newVirtualThreadPerTaskExecutor()creates a new virtual thread per submitted task
, allowing highly concurrent execution.
* Execution Behavior
* cannesFestivalfeatureFilms.stream().limit(25) # Limits the stream to thefirst 25 numbers(1 to
25).
* .forEach(film -> executor.submit(() -> System.out.println(film)))
* Each film is printed inside a virtual thread.
* Virtual threads execute asynchronously, meaning numbers arenot guaranteed to print sequentially.
* Output will contain numbers from 1 to 25, but their order is random due to concurrent execution.
* Possible Output (Random Order)
python-repl
3
1
5
2
4
7
25
* The ordermay differ in each rundue to concurrent execution.
Thus, the correct answer is:"Numbers from 1 to 25 randomly."
References:
* Java SE 21 - Virtual Threads
* Java SE 21 - Executors.newVirtualThreadPerTaskExecutor()

NEW QUESTION # 11
Given:
java
interface A {
default void ma() {
}
}
interface B extends A {
static void mb() {
}
}
interface C extends B {
void ma();
void mc();
}
interface D extends C {
void md();
}
interface E extends D {
default void ma() {
}
default void mb() {
}
default void mc() {
}
}
Which interface can be the target of a lambda expression?

A. A
B. E
C. None of the above
D. B
E. C
F. D

Answer: C

Explanation:
In Java, a lambda expression can be used where a target type is a functional interface. A functional interface is an interface that contains exactly one abstract method. This concept is also known as a Single Abstract Method (SAM) type.
Analyzing each interface:
* Interface A: Contains a single default method ma(). Since default methods are not abstract, A has no abstract methods.
* Interface B: Extends A and adds a static method mb(). Static methods are also not abstract, so B has no abstract methods.
* Interface C: Extends B and declares two abstract methods: ma() (which overrides the default method from A) and mc(). Therefore, C has two abstract methods.
* Interface D: Extends C and adds another abstract method md(). Thus, D has three abstract methods.
* Interface E: Extends D and provides default implementations for ma(), mb(), and mc(). However, it does not provide an implementation for md(), leaving it as the only abstract method in E.
For an interface to be a functional interface, it must have exactly one abstract method. In this case, E has one abstract method (md()), so it qualifies as a functional interface. However, the question asks which interface can be the target of a lambda expression. Since E is a functional interface, it can be the target of a lambda expression.
Therefore, the correct answer is D (E).

NEW QUESTION # 12
Which of the following methods of java.util.function.Predicate aredefault methods?

A. test(T t)
B. isEqual(Object targetRef)
C. negate()
D. or(Predicate<? super T> other)
E. and(Predicate<? super T> other)
F. not(Predicate<? super T> target)

Answer: C,D,E

Explanation:
* Understanding java.util.function.Predicate<T>
* The Predicate<T> interface represents a function thattakes an input and returns a boolean(true or false).
* It is often used for filtering operations in functional programming and streams.
* Analyzing the Methods:
* and(Predicate<? super T> other)#Default method
* Combines two predicates usinglogical AND(&&).
java
Predicate<String> startsWithA = s -> s.startsWith("A");
Predicate<String> hasLength3 = s -> s.length() == 3;
Predicate<String> combined = startsWithA.and(hasLength3);
* #isEqual(Object targetRef)#Static method
* Not a default method, because it doesnot operate on an instance.
java
Predicate<String> isEqualToHello = Predicate.isEqual("Hello");
* negate()#Default method
* Negates a predicate (! operator).
java
Predicate<String> notEmpty = s -> !s.isEmpty();
Predicate<String> isEmpty = notEmpty.negate();
* #not(Predicate<? super T> target)#Static method (introduced in Java 11)
* Not a default method, since it is static.
* or(Predicate<? super T> other)#Default method
* Combines two predicates usinglogical OR(||).
* #test(T t)#Abstract method
* Not a default method, because every predicatemust implement this method.
Thus, the correct answers are:and(Predicate<? super T> other), negate(), or(Predicate<? super T> other) References:
* Java SE 21 - Predicate Interface
* Java SE 21 - Functional Interfaces

NEW QUESTION # 13
Consider the following methods to load an implementation of MyService using ServiceLoader. Which of the methods are correct? (Choose all that apply)

A. MyService service = ServiceLoader.load(MyService.class).iterator().next();
B. MyService service = ServiceLoader.getService(MyService.class);
C. MyService service = ServiceLoader.services(MyService.class).getFirstInstance();
D. MyService service = ServiceLoader.load(MyService.class).findFirst().get();

Answer: A,D

Explanation:
The ServiceLoader class in Java is used to load service providers implementing a given service interface. The following methods are evaluated for their correctness in loading an implementation of MyService:
* A. MyService service = ServiceLoader.load(MyService.class).iterator().next(); This method uses the ServiceLoader.load(MyService.class) to create a ServiceLoader instance for MyService.
Calling iterator().next() retrieves the next available service provider. If no providers are available, a NoSuchElementException will be thrown. This approach is correct but requires handling the potential exception if no providers are found.
* B. MyService service = ServiceLoader.load(MyService.class).findFirst().get(); This method utilizes the findFirst() method introduced in Java 9, which returns an Optional describing the first available service provider. Calling get() on the Optional retrieves the service provider if present; otherwise, a NoSuchElementException is thrown. This approach is correct and provides a more concise way to obtain the first service provider.
* C. MyService service = ServiceLoader.getService(MyService.class);
The ServiceLoader class does not have a method named getService. Therefore, this method is incorrect and will result in a compilation error.
* D. MyService service = ServiceLoader.services(MyService.class).getFirstInstance(); The ServiceLoader class does not have a method named services or getFirstInstance. Therefore, this method is incorrect and will result in a compilation error.
In summary, options A and B are correct methods to load an implementation of MyService using ServiceLoader.

NEW QUESTION # 14
Given:
java
import java.io.*;
class A implements Serializable {
int number = 1;
}
class B implements Serializable {
int number = 2;
}
public class Test {
public static void main(String[] args) throws Exception {
File file = new File("o.ser");
A a = new A();
var oos = new ObjectOutputStream(new FileOutputStream(file));
oos.writeObject(a);
oos.close();
var ois = new ObjectInputStream(new FileInputStream(file));
B b = (B) ois.readObject();
ois.close();
System.out.println(b.number);
}
}
What is the given program's output?

A. NotSerializableException
B. 0
C. 1
D. Compilation fails
E. ClassCastException

Answer: E

Explanation:
In this program, we have two classes, A and B, both implementing the Serializable interface, and a Test class with the main method.
Program Flow:
* Serialization:
* An instance of class A is created and assigned to the variable a.
* An ObjectOutputStream is created to write to the file "o.ser".
* The object a is serialized and written to the file.
* The ObjectOutputStream is closed.
* Deserialization:
* An ObjectInputStream is created to read from the file "o.ser".
* The program attempts to read an object from the file and cast it to an instance of class B.
* The ObjectInputStream is closed.
Analysis:
* Serialization Process:
* The object a is an instance of class A and is serialized into the file "o.ser".
* Deserialization Process:
* When deserializing, the program reads the object from the file and attempts to cast it to class B.
* However, the object in the file is of type A, not B.
* Since A and B are distinct classes with no inheritance relationship, casting an A instance to B is invalid.
Exception Details:
* Attempting to cast an object of type A to type B results in a ClassCastException.
* The exception message would be similar to:
pgsql
Exception in thread "main" java.lang.ClassCastException: class A cannot be cast to class B Conclusion:
The program compiles successfully but throws a ClassCastException at runtime when it attempts to cast the deserialized object to class B.

NEW QUESTION # 15
......

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PASS GUARANTEED 2025 ORACLE HIGH-QUALITY 1Z0-830: NEW JAVA SE 21 DEVELOPER PROFESSIONAL STUDY PLAN

Pass Guaranteed 2025 Oracle High-quality 1z0-830: New Java SE 21 Developer Professional Study Plan